Mixing
A simple modulo arithmetic problem
Clash Royale CLAN TAG#URR8PPP
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I slur `$zmod L$' here to mean the only element of $z+nL: nin mathbbZcap [0,L).$
We are given quantities:
- $a,b, L,$
- $D_1 = (ax mod L) + aw$,
- $D_2 = bx+bw.$
We are also given the fact:
- $D_3 =(ax + bx)in [0,L).$, i.e. $D_3=(D_3mod L)$
From these facts, is it possible to compute the following in terms of the given quantities?
$$D_4 = ax+bx+aw+bw= D_3+aw + bw$$
Here's my try:
beginalign
(D_1+D_2) &= ((ax mod L) + aw + bx+bw) \
&= ((ax mod L)-ax+ax+bx+aw+bw) \
&= ((ax mod L)-ax)+D_4.
endalign
So it is enough to find $(ax mod L)-ax,$ which I am not sure is possible.
algebra-precalculus elementary-number-theory modular-arithmetic arithmetic vector-lattices
asked Jul 29 at 22:33
enthdegree
2,46321235
add a comment |Â
up vote
2
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I slur `$zmod L$' here to mean the only element of $z+nL: nin mathbbZcap [0,L).$
We are given quantities:
- $a,b, L,$
- $D_1 = (ax mod L) + aw$,
- $D_2 = bx+bw.$
We are also given the fact:
- $D_3 =(ax + bx)in [0,L).$, i.e. $D_3=(D_3mod L)$
From these facts, is it possible to compute the following in terms of the given quantities?
$$D_4 = ax+bx+aw+bw= D_3+aw + bw$$
Here's my try:
beginalign
(D_1+D_2) &= ((ax mod L) + aw + bx+bw) \
&= ((ax mod L)-ax+ax+bx+aw+bw) \
&= ((ax mod L)-ax)+D_4.
endalign
So it is enough to find $(ax mod L)-ax,$ which I am not sure is possible.
algebra-precalculus elementary-number-theory modular-arithmetic arithmetic vector-lattices
asked Jul 29 at 22:33
enthdegree
2,46321235
Ok. Comments deleted then. Now I still don't really understand how you want to derive something non-trivial from the three given equalities: the first two define $D_1$ and $D_2$, they don't really carry information. The third one does carry the information that $0leq (a+b)x <L$. That's basically all you have. Finally, what do you mean by " is it possible to compute..."? Do you mean compute "the" representative of that modulo $L$? And how is that question related to $D_1$ and $D_2$?
– Arnaud Mortier
Jul 29 at 22:59
No, I mean the quantity outright. That is, if $D_3=(a+b)cdot x in [0,L)$ then I want $D_3+(a+b)w in mathbbR$
– enthdegree
Jul 29 at 23:02
So to be clear, you want an answer in terms of $D_1$ and $D_2$?
– é«˜ç”°èˆª
Jul 29 at 23:12
Yes. $$
– enthdegree
Jul 29 at 23:14
You don't know $x$ and $w$ but you know $a$ and $b$ is that right?
– Arnaud Mortier
Jul 31 at 22:02
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I slur `$zmod L$' here to mean the only element of $z+nL: nin mathbbZcap [0,L).$
We are given quantities:
- $a,b, L,$
- $D_1 = (ax mod L) + aw$,
- $D_2 = bx+bw.$
We are also given the fact:
- $D_3 =(ax + bx)in [0,L).$, i.e. $D_3=(D_3mod L)$
From these facts, is it possible to compute the following in terms of the given quantities?
$$D_4 = ax+bx+aw+bw= D_3+aw + bw$$
Here's my try:
beginalign
(D_1+D_2) &= ((ax mod L) + aw + bx+bw) \
&= ((ax mod L)-ax+ax+bx+aw+bw) \
&= ((ax mod L)-ax)+D_4.
endalign
So it is enough to find $(ax mod L)-ax,$ which I am not sure is possible.
algebra-precalculus elementary-number-theory modular-arithmetic arithmetic vector-lattices
asked Jul 29 at 22:33
enthdegree
2,46321235
I slur `$zmod L$' here to mean the only element of $z+nL: nin mathbbZcap [0,L).$
We are given quantities:
- $a,b, L,$
- $D_1 = (ax mod L) + aw$,
- $D_2 = bx+bw.$
We are also given the fact:
- $D_3 =(ax + bx)in [0,L).$, i.e. $D_3=(D_3mod L)$
From these facts, is it possible to compute the following in terms of the given quantities?
$$D_4 = ax+bx+aw+bw= D_3+aw + bw$$
Here's my try:
beginalign
(D_1+D_2) &= ((ax mod L) + aw + bx+bw) \
&= ((ax mod L)-ax+ax+bx+aw+bw) \
&= ((ax mod L)-ax)+D_4.
endalign
So it is enough to find $(ax mod L)-ax,$ which I am not sure is possible.
algebra-precalculus elementary-number-theory modular-arithmetic arithmetic vector-lattices
asked Jul 29 at 22:33
enthdegree
2,46321235
edited Jul 29 at 23:22
asked Jul 29 at 22:33
enthdegree
2,46321235
asked Jul 29 at 22:33
enthdegree
2,46321235
asked Jul 29 at 22:33
enthdegree
2,46321235
2,46321235
Ok. Comments deleted then. Now I still don't really understand how you want to derive something non-trivial from the three given equalities: the first two define $D_1$ and $D_2$, they don't really carry information. The third one does carry the information that $0leq (a+b)x <L$. That's basically all you have. Finally, what do you mean by " is it possible to compute..."? Do you mean compute "the" representative of that modulo $L$? And how is that question related to $D_1$ and $D_2$?
– Arnaud Mortier
Jul 29 at 22:59
No, I mean the quantity outright. That is, if $D_3=(a+b)cdot x in [0,L)$ then I want $D_3+(a+b)w in mathbbR$
– enthdegree
Jul 29 at 23:02
So to be clear, you want an answer in terms of $D_1$ and $D_2$?
– é«˜ç”°èˆª
Jul 29 at 23:12
Yes. $$
– enthdegree
Jul 29 at 23:14
You don't know $x$ and $w$ but you know $a$ and $b$ is that right?
– Arnaud Mortier
Jul 31 at 22:02
add a comment |Â
Ok. Comments deleted then. Now I still don't really understand how you want to derive something non-trivial from the three given equalities: the first two define $D_1$ and $D_2$, they don't really carry information. The third one does carry the information that $0leq (a+b)x <L$. That's basically all you have. Finally, what do you mean by " is it possible to compute..."? Do you mean compute "the" representative of that modulo $L$? And how is that question related to $D_1$ and $D_2$?
– Arnaud Mortier
Jul 29 at 22:59
No, I mean the quantity outright. That is, if $D_3=(a+b)cdot x in [0,L)$ then I want $D_3+(a+b)w in mathbbR$
– enthdegree
Jul 29 at 23:02
So to be clear, you want an answer in terms of $D_1$ and $D_2$?
– é«˜ç”°èˆª
Jul 29 at 23:12
Yes. $$
– enthdegree
Jul 29 at 23:14
You don't know $x$ and $w$ but you know $a$ and $b$ is that right?
– Arnaud Mortier
Jul 31 at 22:02
Ok. Comments deleted then. Now I still don't really understand how you want to derive something non-trivial from the three given equalities: the first two define $D_1$ and $D_2$, they don't really carry information. The third one does carry the information that $0leq (a+b)x <L$. That's basically all you have. Finally, what do you mean by " is it possible to compute..."? Do you mean compute "the" representative of that modulo $L$? And how is that question related to $D_1$ and $D_2$?
– Arnaud Mortier
Jul 29 at 22:59
Ok. Comments deleted then. Now I still don't really understand how you want to derive something non-trivial from the three given equalities: the first two define $D_1$ and $D_2$, they don't really carry information. The third one does carry the information that $0leq (a+b)x <L$. That's basically all you have. Finally, what do you mean by " is it possible to compute..."? Do you mean compute "the" representative of that modulo $L$? And how is that question related to $D_1$ and $D_2$?
– Arnaud Mortier
Jul 29 at 22:59
No, I mean the quantity outright. That is, if $D_3=(a+b)cdot x in [0,L)$ then I want $D_3+(a+b)w in mathbbR$
– enthdegree
Jul 29 at 23:02
No, I mean the quantity outright. That is, if $D_3=(a+b)cdot x in [0,L)$ then I want $D_3+(a+b)w in mathbbR$
– enthdegree
Jul 29 at 23:02
So to be clear, you want an answer in terms of $D_1$ and $D_2$?
– é«˜ç”°èˆª
Jul 29 at 23:12
So to be clear, you want an answer in terms of $D_1$ and $D_2$?
– é«˜ç”°èˆª
Jul 29 at 23:12
Yes. $$
– enthdegree
Jul 29 at 23:14
Yes. $$
– enthdegree
Jul 29 at 23:14
You don't know $x$ and $w$ but you know $a$ and $b$ is that right?
– Arnaud Mortier
Jul 31 at 22:02
You don't know $x$ and $w$ but you know $a$ and $b$ is that right?
– Arnaud Mortier
Jul 31 at 22:02
add a comment |Â
1 Answer
1
active
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votes
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0
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accepted
The difficulty of that problem is that there is a lot of useless data. All you need is that you know $a$, $b$, and $D_2=bx+bw$.
$$D_4=D_2left(1+frac abright)$$
answered Jul 31 at 22:05
Arnaud Mortier
18.2k21958
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The difficulty of that problem is that there is a lot of useless data. All you need is that you know $a$, $b$, and $D_2=bx+bw$.
$$D_4=D_2left(1+frac abright)$$
answered Jul 31 at 22:05
Arnaud Mortier
18.2k21958
add a comment |Â
up vote
0
down vote
accepted
The difficulty of that problem is that there is a lot of useless data. All you need is that you know $a$, $b$, and $D_2=bx+bw$.
$$D_4=D_2left(1+frac abright)$$
answered Jul 31 at 22:05
Arnaud Mortier
18.2k21958
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The difficulty of that problem is that there is a lot of useless data. All you need is that you know $a$, $b$, and $D_2=bx+bw$.
$$D_4=D_2left(1+frac abright)$$
answered Jul 31 at 22:05
Arnaud Mortier
18.2k21958
The difficulty of that problem is that there is a lot of useless data. All you need is that you know $a$, $b$, and $D_2=bx+bw$.
$$D_4=D_2left(1+frac abright)$$
answered Jul 31 at 22:05
Arnaud Mortier
18.2k21958
answered Jul 31 at 22:05
Arnaud Mortier
18.2k21958
answered Jul 31 at 22:05
Arnaud Mortier
18.2k21958
answered Jul 31 at 22:05
Arnaud Mortier
18.2k21958
18.2k21958
add a comment |Â
add a comment |Â
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Ok. Comments deleted then. Now I still don't really understand how you want to derive something non-trivial from the three given equalities: the first two define $D_1$ and $D_2$, they don't really carry information. The third one does carry the information that $0leq (a+b)x <L$. That's basically all you have. Finally, what do you mean by " is it possible to compute..."? Do you mean compute "the" representative of that modulo $L$? And how is that question related to $D_1$ and $D_2$?
– Arnaud Mortier
Jul 29 at 22:59
No, I mean the quantity outright. That is, if $D_3=(a+b)cdot x in [0,L)$ then I want $D_3+(a+b)w in mathbbR$
– enthdegree
Jul 29 at 23:02
So to be clear, you want an answer in terms of $D_1$ and $D_2$?
– é«˜ç”°èˆª
Jul 29 at 23:12
Yes. $$
– enthdegree
Jul 29 at 23:14
You don't know $x$ and $w$ but you know $a$ and $b$ is that right?
– Arnaud Mortier
Jul 31 at 22:02